Moment of inertia of rod about an axis inclined

[Raghav Gupta]

No, not squash it to make it parallel to the axis, squash it in a direction parallel to the axis (making it perpendicular to the axis). Each bit of the rod should stay the same distance from the axis in the squashing process.

But the angle between axis and rod would change?

 

[haruspex]

But the angle between axis and rod would change?

Sure, and that's the point. The resulting shorter rod would have the same MoI about the axis as the original. So all you have to do is work out how long the rod is now and apply the usual formula.

 

[Raghav Gupta]

Sure, and that's the point. The resulting shorter rod would have the same MoI about the axis as the original. So all you have to do is work out how long the rod is now and apply the usual formula.

I may be asking too much but I am not able to imagine why the length would change of rod? Angle changing I am able to imagine.

 

[haruspex]

I may be asking too much but I am not able to imagine why the length would change of rod? Angle changing I am able to imagine.

Let L be a line perpendicular to the axis, meeting it where the rod does.
Imagine slicing the rod into short sections, the cuts being parallel to rotation the axis. Now slide them, cut surface sliding on cut surface, so that all the pieces now lie on line L. Ok, it's a bit jagged along the sides of the rod, but by making the sections very short we can reduce the jaggedness as much as we like, so in effect we now have a shorter but fatter rod. Each little piece is the same distance from the axis as it was before, so the MoI has not changed.

 

[Raghav Gupta]

Let L be a line perpendicular to the axis, meeting it where the rod does.
Imagine slicing the rod into short sections, the cuts being parallel to rotation the axis. Now slide them, cut surface sliding on cut surface, so that all the pieces now lie on line L. Ok, it's a bit jagged along the sides of the rod, but by making the sections very short we can reduce the jaggedness as much as we like, so in effect we now have a shorter but fatter rod. Each little piece is the same distance from the axis as it was before, so the MoI has not changed.

Okay, got it.
The new length would be Lsinθ and we know the formula for axis perpendicular to rod as ML²/12. So now formula would be ML²sin²θ/12 .

 

[haruspex]

Okay, got it.
The new length would be Lsinθ and we know the formula for axis perpendicular to rod as ML²/12. So now formula would be ML²sin²θ/12 .

According to your diagram and explanation at the start of the thread, it would be cos, not sin. But I wonder if you drew the diagram correctly. Shouldn't theta be the angle between the rod and the axis?

 

[Raghav Gupta]

According to your diagram and explanation at the start of the thread, it would be cos, not sin. But I wonder if you drew the diagram correctly. Shouldn't theta be the angle between the rod and the axis?

No, I corrected it in post 4 or 5 around. Yes theta is angle between rod and axis.
Thanks to you as well Haruspex.

 

[BvU]

I hate to interfere with all this lovely integration, but it really isn't necessary.
The contribution of a mass element to the MoI depends only on its distance from the axis. So you can squash the object parallel to the axis, as long as you preserve masses. In this case, you will have a shorter, fatter (or higher density) rod. The mass will not change. What will the new length be?

Haru is absoltely right. So if you have something to start with, this is a nice tool to take the most effective approach. In other cases, it's good to have something as basic as $$
I = \int r^2 dm$$ to start from.

And you can also see right through this ( I didn't until now) and realize that dm can be ... dx but also ... dy. And the ... dy work has already been done (if you know the formula or have a table at hand).